2= (>>60) (z<80) 5. A The ages of the thousands of residents of a retirement community are normally distributed with a mean of 70 and a standard deviation of 4 years. a. What proportion of this population is between 60 and 80? X.4 2= 2=60-70 = -2.5 = 6.21 4 2-80-76 J = 2.5=6.21 1-,99379 -3 = 6.21% +6.2 12.24% is bett 60- Year: b. If one sample of 45 residents is chosen at random, what is the probability that the sample mean age will be between 68.5 and 71.75? 68.5-45 2= 25.87

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**Transcription for Educational Website**

---

**Problem 5**

A: The ages of the thousands of residents of a retirement community are normally distributed with a mean of 70 and a standard deviation of 4 years.

**a. What proportion of this population is between 60 and 80?**

\[
Z = \frac{X - \mu}{\sigma}
\]

- For age 60:
  \[
  Z = \frac{60 - 70}{4} = -2.5 \approx -6.21
  \]

- For age 80:
  \[
  Z = \frac{80 - 70}{4} = 2.5 \approx 6.21
  \]

Calculation: 
1 - (Probability of Z < -6.21 + Probability of Z > 6.21) = 0.99379

Proportion = 6.21% + 6.21% = 12.24%

**Conclusion:** 12.24% of the population is between 60 and 80 years old.

---

**b. If one sample of 45 residents is chosen at random, what is the probability that the sample mean age will be between 68.5 and 71.75?**

\[
Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}}
\]

- For age 68.5:
  \[
  Z = \frac{68.5 - 70}{4/\sqrt{45}} = -5.87
  \]

- For age 71.75:
  \[
  Z = \frac{71.75 - 70}{4/\sqrt{45}} = 6.6
  \]

Diagrams and detailed calculations show that this configuration results in a probability of a sample mean being in this range, with students marked -3 for incorrect graph plotting and -4 for calculation errors. 

**Note:** Exact Z scores, such as -6.21, indicate more precise computations and rounding in context.
Transcribed Image Text:**Transcription for Educational Website** --- **Problem 5** A: The ages of the thousands of residents of a retirement community are normally distributed with a mean of 70 and a standard deviation of 4 years. **a. What proportion of this population is between 60 and 80?** \[ Z = \frac{X - \mu}{\sigma} \] - For age 60: \[ Z = \frac{60 - 70}{4} = -2.5 \approx -6.21 \] - For age 80: \[ Z = \frac{80 - 70}{4} = 2.5 \approx 6.21 \] Calculation: 1 - (Probability of Z < -6.21 + Probability of Z > 6.21) = 0.99379 Proportion = 6.21% + 6.21% = 12.24% **Conclusion:** 12.24% of the population is between 60 and 80 years old. --- **b. If one sample of 45 residents is chosen at random, what is the probability that the sample mean age will be between 68.5 and 71.75?** \[ Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}} \] - For age 68.5: \[ Z = \frac{68.5 - 70}{4/\sqrt{45}} = -5.87 \] - For age 71.75: \[ Z = \frac{71.75 - 70}{4/\sqrt{45}} = 6.6 \] Diagrams and detailed calculations show that this configuration results in a probability of a sample mean being in this range, with students marked -3 for incorrect graph plotting and -4 for calculation errors. **Note:** Exact Z scores, such as -6.21, indicate more precise computations and rounding in context.
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